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Suppose,there are six molecules(indistinguishable) constantly jostling around in a box. The central configuration is when there are $3$ molecules in either halves of the box and having probability $31.3$%. However,

it is surprising that there is probability,however small, of finding all the molecules in the corner half of the box. - Resnick,Halliday,Walker.

Doesn't it break the Second law of thermodynamics? The configuration where the molecules are confined in a corner isn't possible. But math is showing otherwise. They have probability $1.6$%. How can it be possible??

[Even Planck also hesitated and at first rejected the underlying assumption of Boltzmann's approach which allows the Second law to be violated momentarily during $\text{fluctuations}^1$.

$1$"A small probability exists that all the molecules of a system of confined gas might appear for an instant in just one corner of the container. This is called energy fluctuation."- Boltzmann]

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  • $\begingroup$ The second law of thermodynamics is the definition of temperature. How do you see it being violated here? $\endgroup$ – CuriousOne Jan 2 '15 at 5:09
  • $\begingroup$ you should edit the probability numbers. Probability is always less than 1. you must mean % $\endgroup$ – anna v Jan 2 '15 at 5:32
  • $\begingroup$ @anna v: I had written %, but it didn't appear:-( $\endgroup$ – user36790 Jan 2 '15 at 5:35
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Here are the laws of thermodynamics and the second one:

Second law of thermodynamics: In a natural thermodynamic process, the sum of the entropies of the participating thermodynamic systems increases. Equivalently, perpetual motion machines of the second kind are impossible.

You are discussing six molecules. Thermodynamics is a theory formulated to fit observations for large numbers of particles , of the order of the Avogadro number ~10^23. It is a very successful theory used in many aspects of engineering, but that does not mean that it scales down to small numbers . So there is no problem with your example.

The statement you quote

"A small probability exists that all the molecules of a system of confined gas might appear for an instant in just one corner of the container. This is called energy fluctuation."- Boltzmann]

Is a visualized extension from small numbers, adding more and more molecules will make the probabilities smaller and smaller, and because of the 10^23 extremely improbable.

This is the study of thermodynamics as it emerges from statistical mechanics, where it is shown that entropy is not a "law" in the same sense as energy conservation is a law, but can be derived from the statistics of many bodies.

entropy

In statistical mechanics, Boltzmann's equation is a probability equation relating the entropy S of an ideal gas to the quantity W, which is the number of microstates corresponding to a given macrostate,

Each individual microstate is equally improbable as getting all the molecules on one side of the box, it is the sheer numbers of the average mix that define entropy at the thermodynamic framework, giving probabilities of order of 10^-12. The link may help.

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  • $\begingroup$ Just a suggestion: if you want to teach someone about the basics of thermodynamics, don't start with Boltzmann. Start with Clausius: "Heat can never pass from a colder to a warmer body without some other change, connected therewith, occurring at the same time.". For one thing it's way more instructive, for another it takes all the air out of false ideas about "entropy" (everybody and their grandma think they look smart when they use the word entropy, and almost everybody and their grandma keep applying it wrong). $\endgroup$ – CuriousOne Jan 2 '15 at 6:01
  • $\begingroup$ @CuriousOne I am just answering the question, not teaching thermodynamics, I am explaining the connection of probabilities from small numbers to large numbers and thermodynamics. $\endgroup$ – anna v Jan 2 '15 at 7:33
  • $\begingroup$ The OP's problem has nothing to do with the second law, since it's completely independent of temperature and heat flow. One doesn't need Boltzmann where it doesn't apply. $\endgroup$ – CuriousOne Jan 2 '15 at 9:16
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Let's consider your six particles as members of a non-interacting ideal gas.

  • The probability that the first particle is somewhere in the left half of the box is one-half.
  • The probability that the second particle is somewhere in the left half of the box is one-half.

This tells that the probability of having all six particles in the left half of the box is $2^{-6} \approx 1.6\%$, as you say: that is only one configuration out of the $2^6 = 64$ possibilities. By comparison there are "six choose three," or $$ {6 \choose 3} = \frac{6!}{3!\,3!} = \frac{6\cdot5\cdot4}{3\cdot2\cdot1} = 20 $$ possible configurations where you have three particles on the left side. (Six could be the first one over, five remain to be the second, four for the third, but there are $3!=6$ ways to order them, and we don't want to over-count.) That means that you'll find exactly three particles on the left side about $20/64 \approx 31\%$ of the time, as you say. In fact you'll have $3\pm\sqrt3$ particles on the left side somewhere between 70% and 97% of the time, depending on whether you count the 5/1 state or not. Here's a little histogram of how much time your box will be in each state:

particles on left side | number of states
 0 | .1
 1 | ......6
 2 | ...............15
 3 | ....................20
 4 | ...............15
 5 | ......6
 6 | .1

Now, consider what changes if you go to twelve particles in your "gas." You'll still have a $2^{-6}$ chance that your original six will all be on the left side; but whenever that happens, you'll have a 97% chance that at least one of the six new particles is on the "empty" right side, so it's no longer fair to call it empty. Look at the histogram for a twelve-particle box:

 0 | .1
 1 | ............12
 2 | ..................................................................66
 3 | ............................................................................................................................................................................................................................220
 4 | ...............................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................495
 5 | ........................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................792
 6 | ............................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................924
 7 | ........................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................792
 8 | ...............................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................495
 9 | ............................................................................................................................................................................................................................220
10 | ..................................................................66
11 | ............12
12 | .1

See, in order to make that $2^{-12}$ chance of an empty side visible, it's necessary to make the rest of the histogram completely useless. Let's see what happens if we try to squeeze it onto our screen here:

 0 | 00%
 1 | 00%
 2 | ....02%
 3 | ................05%
 4 | ....................................12%
 5 | ..........................................................19%
 6 | ...................................................................23%
 7 | ..........................................................19%
 8 | ....................................12%
 9 | ................05%
10 | ....02%
11 | 00%
12 | 00%

We find that that there are between four and eight particles (that's $6\pm\sqrt6$) on each side about 85% of the time. Double the population in the box again:

 4 | 00%
 5 | .00%
 6 | ...01%
 7 | ........02%
 8 | .................04%
 9 | ...............................08%
10 | ..............................................12%
11 | ...........................................................15%
12 | ................................................................16%
13 | ...........................................................15%
14 | ..............................................12%
15 | ...............................08%
16 | .................04%
17 | ........02%
18 | ...01%
19 | .00%
20 | 00%

and again, to 48 particles

13 | 00%
14 | .00%
15 | ..00%
16 | ....01%
17 | .........02%
18 | ...............03%
19 | ........................04%
20 | ...................................06%
21 | ...............................................08%
22 | ..........................................................10%
23 | .................................................................11%
24 | ....................................................................11%
25 | .................................................................11%
26 | ..........................................................10%
27 | ...............................................08%
28 | ...................................06%
29 | ........................04%
30 | ...............03%
31 | .........02%
32 | ....01%
33 | ..00%
34 | .00%
35 | 00%

Notice that for these two I haven't copied over most of the rows where the probability is too low to print a dot. This is honestly hidden data: those zeros really are zeros. The shape of the curve is called the binomial distribution and it becomes quite similar to the better-known bell curve for many particles. If you have $2N$ particles in your box, then

  • about 85% of the time you'll find $N\pm\sqrt N$ on the left side;
  • about 99.5% of the time you'll find $N\pm2\sqrt N$ on the left side;
  • about 99.998% of the time you'll find $N\pm3\sqrt N$ on the left side.

If you had only 1000 particles in your box, there are only about five configurations out of every $10^{10}$ that a split between the two sides more extreme than 400/600; if your box took a millisecond to cross, like a typical meter-sized box of gas at room temperature, you'd see two-to-one density difference between the sides about once a month. For a millisecond.

The thing that breaks your intuition about statistical mechanics is the sheer size of the numbers involved. I gave you some examples for up to 1000 particles because I happen to be using computer software which tolerates numbers up to $2^{1024}$. But a liter of gas at STP contains something like $10^{23}$ particles, which have $2^{10^{23}}$ ways to partition themselves between the two halves of a box. The natural density fluctuations due to particles randomly moving between the two sides of the box are at the level of $\sqrt{10^{23}} = 10^{11.5}$ particles — changes which begin in the eleventh digit of the number. For example, if I expected to find 100 000 000 000 000 000 000 000 atoms on the left side of the box on average, about 7.5% of the time I'd randomly find more than 100 000 000 000 030 000 000 000 atoms, and about 0.001% of the time I'd find more than 100 000 000 000 100 000 000 000 atoms. Totally unmeasurable differences. The fluctuations are made small by the statistical size of the ensemble.

It's not impossible for a mole of gas to briefly, randomly find itself in one half of its container. But it's a safe bet that no such extreme fluctuation has occurred for any such volume in the observable universe since the Big Bang. Your intuition has failed you because $6 \neq 10^{23}$, not even for extremely large values of six.

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